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If 10.0g of powdered iron is heated with 10.0g of sulfur in an open crucible, what is the mass of iron (II) sulfide that is formed? What mass of excess reactant is leftover? The reaction is as follows: Fe+S8–> FeS

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Answer:

See Explanation

Explanation:

                     8Fe        +        S₈                =>        8FeS

Given:            10g                  10g

moles      10g/56g/mol     10g/256g/mol

                = 0.179mol Fe   = 0.039mol S₈

Reduce => divide mole values by respective coefficients; smaller value is Limiting Reactant.

                 0.179/8 = 0.022      0.039/1 = 0.039  

       => Fe is limiting reactant

                       8Fe        +               S₈                   =>         8FeS    

Given:     10g/56g/mol          10g/256g/mol

               = 0.179mol              = 0.039 mol                  0.179mol FeS produced

                                          1/8(0.179)mol S₈ used        (coefficients are equal,

                                          = 0.022 mol S₈ used       => moles Fe = moles FeS)

                                         = (0.039 - 0.022)mol S₈     = 0.179mol FeS

                                         remains in excess              =(0.179mol)(88g/mol)

                                         = 0.0166 mol S₈ (excess)  = 15.8 g FeS  

                                         = (0.0166mol)(256g/mol)           (Theoretical Yield)

                                         = 4.26g S₈ in excess

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